The tangent line to the graph of function $h$ at the point $(3,8)$ passes through the point $(-1,0)$. Find $h'(3)$. $h'(3)=$
Answer: The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $h'(3)$ gives the slope of the tangent line to the graph of $h$ where $x=3$, which is the point $(3,8)$. We know this line passes through $(3,8)$, and we are also given that it passes through $(-1,0)$. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{8-0}{3-(-1)} \\\\ &=\dfrac{8}{4} \\\\ &=2 \end{aligned}$ In conclusion, $h'(3)=2$.